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From Wikipedia:

Let $R$ be a ring of subsets of a set $Ω$ and $μ: R → [0, + ∞]$ be a $\sigma$-additive pre-measure on a $R$.

The Carathéodory's extension theorem states that there exists a measure $μ′: σ(R) → [0, + ∞]$ such that $μ′$ is an extension of $μ$.

I was wondering if an extension exists when $R$ is a $\pi$ system?

The reason for considering a $\pi$ system is because a measure on a sigma algebra generated by a $\pi$ system is uniquely determined by its restriction on the $\pi$ system, if the measure is $\sigma$-finite wrt the $\pi$ system. I want to know about the existence of extension instead.

Thanks an regards!

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Let $\Omega = (0,1)$ and let $R = \{(0,t) : 0 < t < 1\}$ be the collection of all open intervals whose left endpoint is 0. $R$ is certainly a $\pi$-system, and $\sigma(R)$ is the Borel $\sigma$-algebra. Define $\mu : R \to [0, +\infty]$ by $\mu(A) = 1$ for all $A \in R$. This is trivially $\sigma$-additive because $R$ contains no disjoint sets! But $\mu$ cannot extend to a $\sigma$-additive measure $\mu'$ on $\sigma(R)$. (If $\mu'((0,t)) = 1$ for all $t$ then by continuity from above $\mu(\emptyset) = 1$ which is absurd.)

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Thanks! Is $\mu$ also $\sigma$-finite wrt $R$, because there exists $A_n := (0, \frac{n-1}{n}), n \in\mathbb{N}$ that satisfy $\cup_n A_n = \Omega$ and $\mu(A_n)=1< \infty$? –  Tim Jan 14 '13 at 5:27
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